PHY 140A: Solid State Physics. Solution to Homework #7

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1 PHY 14A: Solid State Physics Solution to Homework #7 Xun Jia 1 December 5, jiaxun@physics.ucla.edu

2 Fall 26 Physics 14A c Xun Jia (December 5, 26) Problem #1 Static magnetoconductivity tensor. For the drift velocity theory of (51), show that the static current density can be written in matrix form as: j x 1 ω j y σ c τ E x = ω 1 + (ω j c τ) 2 c τ 1 E y (1) z 1 + (ω c τ) 2 E z In the high magnetic field limit of ω c τ 1, show that σ yx = nec/b = σ xy (2) In this limit σ xx =, to order 1/ω c τ. The quantity σ yx is called the Hall conductivity. From Eqn. (51) in Kittel, in the static case, the velocity is not varying with time, so dv i /dt = for i = x, y, z, therefore, in the matrix form, we have: m/τ eb/c v x E x eb/c m/τ v y = e E y () m/τ v z E z solve this equation by inverting the coefficient matrix, and let ω c = eb/mc, we get: v x 1 ω v y eτ c τ E x = ω m[1 + (ω v c τ) 2 c τ 1 E y (4) ] z 1 + (ω c τ) 2 E z since j = nev and σ = ne 2 τ/m, it follows that: j x v x j y j z = ne v y v z = σ 1 + (ω c τ) 2 1 ω c τ ω c τ (ω c τ) 2 E x E y E z (5) and: In the high magnetic field limit, ω c τ 1, then: σ yx = σ xy = σ xx = thus σ xx =, to order 1/ω c τ. σ ω c τ 1 + (ω c τ) σ ω c τ 2 (ω c τ) = σ 2 ω c τ = nec B σ 1 + (ω c τ) σ ( ) 1 2 (ω c τ) = + σ (7) 2 ω c τ (ω c τ) 2 1 (6)

3 Fall 26 Physics 14A c Xun Jia (December 5, 26) Problem #2 Kinetic energy of electron gas. Show that the kinetic energy of a three dimensional gas of N free electrons at K is U = 5 Nɛ F (8) For the Fermi-Dirac distribution at K, we have: { 1 f(ɛ) = 1 : ɛ < ɛf e (ɛ ɛ F )/k BT + 1 = (9) T = : ɛ > ɛ F and the density of states of electrons in -D is : D(ɛ) = V dk 4πk2 (2π) dɛ = V 4π 2 ( ) /2 2m h 2 ɛ 1/2 (1) since for free electrons, the kinetic energy equals to the total energy ɛ, the average kinetic energy per electron is simply: ɛ = dɛd(ɛ)f(ɛ)ɛ = dɛd(ɛ)f(ɛ) ɛ F dɛd(ɛ)ɛ = 5 ɛ F (11) dɛd(ɛ) ɛ F therefore the kinetic energy of total N electrons is: U = N ɛ = 5 Nɛ F (12) Problem # Chemical potential in two dimensions. Show that the chemical potential of a Fermi gas in two dimensions is given by: µ(t ) = k B T ln[exp(πn h 2 /mk B T ) 1] (1) for n electrons per unit area. Note: the density of orbitals of a free electron gas in two dimensions is independent of energy: D(ɛ) = m/, per unit area of specimen. At finite temperature T, the chemical potential is determined by the condition: n = dɛd(ɛ)f(ɛ) = dɛ m 1 e (ɛ µ)/k BT + 1 (14) 2

4 Fall 26 Physics 14A c Xun Jia (December 5, 26) the right hand side can be computed explicitly as: dɛ m 1 e (ɛ µ)/k BT + 1 = m = mk BT = mk BT = mk BT e (ɛ µ)/k BT dɛ 1 + e (ɛ µ)/k BT so from above two equations, we can solve for µ as: [ ( πn h 2 ) µ(t ) = k B T ln exp mk B T d(e (ɛ µ)/kbt ) 1 + e (ɛ µ)/k BT [ ( ln 1 + exp ɛ µ )] k B T [ ( )] µ ln 1 + exp k B T ] 1 ɛ= (15) (16) Problem #4 Fermi gases in astrophysics. (a). Given M = 2 1 g for the mass of the Sun, estimate the number of electrons in the Sun. In a white dwarf star this number of electrons may be ionized and contained in a sphere of radius cm; find the Fermi energy of the electrons in electron volts. (b). The energy of an electron in the relativistic limit ɛ mc 2 is related to the wavevector as ɛ pc = hkc. Show that the Fermi energy in this limit is ɛ F hc(n/v ) 1/, roughly. (c). If the above number of electrons were contained within a pulsar of radius 1km, show that the Fermi energy would be 1 8 ev. This explains why pulsars are believed to be composed largely of neutrons rather than of protons and electrons, for the energy release in the reaction n p + e is only ev, which is not large enough to enable many electrons to form a Fermi sea. The neutron decay proceeds only until the electron concentration builds up enough to create a Fermi level of ev, at which point the neutron, proton, and electron concentrations are in equilibrium. (a). The Sun is mainly composed by hydrogen atoms, whose mass is approximately equal to the mass of proton m p. Since each hydrogen atom contains one electron, the number of electrons in the Sun is: N = M m p = (17)

5 Fall 26 Physics 14A c Xun Jia (December 5, 26) if this number of electrons is contained in a white dwarf star with a radius R wd = cm, the Fermi energy is: 2/ ( ɛ F = h2 π 2 N ) 2/ = h2 N 2m V 2m π2 4πRwd (18) = J = ev (b). In the relativistic case with a dispersion relation of ɛ pc = hkc, the density of states of electrons is: D(ɛ) = 2 V dk 4πk2 (2π) dɛ = V ɛ 2 π 2 h (19) c then the Fermi energy is determined by the condition: which gives: ɛ F = ɛ F D(ɛ)dɛ = V ɛ F π 2 h c (2) ( ɛ F = hc π 2 N ) 1/ ( ) 1/ N hc (21) V V (c). If above number of electrons is contained in a pulsar of radius R p = 1km, the Fermi would be: 1/ ( ) 1/ N ɛ F = hc = hc N V 4πRp = J = ev (22) Problem #5 Liquid He. The atom He has spin 1 and is a fermion. The density of liquid 2 He is.81g/cm near absolute zero. Calculate the Fermi energy ɛ F and the Fermi temperature T F. The mass of a single He is about: m He = a.m.u. = kg = kg (2) therefore the number density of He is n = ρ m He = m (24) 4

6 Fall 26 Physics 14A c Xun Jia (December 5, 26) so the Fermi energy is: ɛ F = h2 2m He ( π 2 N ) 2/ = ɛ F = h2 (π 2 n) 2/ V 2m He = J = ev (25) and the Fermi temperature is: T F = ɛ F k B = 4.88K (26) 5